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If K is a subgroup of G of order p^k, show that K is subgroup of H

  1. Apr 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Let |G| = (p^n)m where p is prime and gcd(p,m) = 1. Suppose that H is a normal subgroup of G of order p^n. If K is a subgroup of G of order p^k, show that K is subgroup of H.



    2. Relevant equations



    3. The attempt at a solution
    Okay, I wonder if there is more I need to do, or if I need to prove they are finite. I feel like I am missing something...but here is what I got
    p^k has to be less than p^n because if p^k was bigger than p^n then p^k would not divide the order of G because p and m are relatively prime and K could not be a subgroup of G. The order of a subgroup must divide the order of the group.

    Both H and K are subgroups of G, they both are closed under the same operation as G, and because n>k, p^k divides p^n and thus because K is closed under the operation of H and K's order divides the order of H, K must be a subgroup of H.

    Thanks
     
  2. jcsd
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